Editing Positive operator (section)

== Cauchy–Schwarz inequality ==
{{Main|Cauchy–Schwarz inequality}}
Take  the inner product <math>\langle \cdot, \cdot \rangle</math> to be [[Antilinear map|anti-linear]] on the ''first'' argument and linear on the second and suppose that <math>A </math> is positive and symmetric, the latter meaning that   <math>  \langle Ax,y \rangle= \langle x,Ay \rangle </math>.
Then  the non negativity of 
:<math>
\begin{align}
 \langle A(\lambda x+\mu y),\lambda x+\mu y \rangle
=|\lambda|^2 \langle Ax,x \rangle + \lambda^* \mu \langle Ax,y \rangle+ \lambda \mu^* \langle Ay,x \rangle + |\mu|^2 \langle Ay,y \rangle \\[1mm]
= |\lambda|^2 \langle Ax,x \rangle + \lambda^* \mu \langle Ax,y \rangle+ \lambda \mu^* (\langle Ax,y \rangle)^* + |\mu|^2 \langle Ay,y \rangle
\end{align}
</math>
for all complex <math>\lambda </math> and <math> \mu </math> shows that 
:<math>\left|\langle Ax,y\rangle \right|^2 \leq \langle Ax,x\rangle \langle Ay,y\rangle.</math>
It follows that <math>\mathop{\text{Im}}A \perp \mathop{\text{Ker}}A.</math> If <math>A</math> is defined everywhere, and <math>\langle Ax,x\rangle = 0,</math> then <math>Ax = 0.</math>